Acceleration of a downhill skier

  • Thread starter Thread starter aron silvester
  • Start date Start date
  • Tags Tags
    Acceleration
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 2K views
aron silvester

Homework Statement


I understand in my head that Wx = mgsin(27) and Wy = -mgcos(27). Though when I tried solving for both Wx and Wy, their signs turned out to be opposite. I've provided my work leading up to Wx = -mgsin(27) and Wy = mgcos(27). Maybe I interpreted the signs of x and y component of W?
IMG_1797.jpeg


Homework Equations

The Attempt at a Solution


IMG_1798.jpeg
[/B]
 

Attachments

  • IMG_1797.jpeg
    IMG_1797.jpeg
    65 KB · Views: 1,093
  • IMG_1798.jpeg
    IMG_1798.jpeg
    54.6 KB · Views: 778
Physics news on Phys.org
The weight vector should be denoted ##\vec w##, not ##-\vec w##. The vector ##-\vec w## would point in the opposite direction of the weight.

Also, in you diagram, you drew ##\vec w_x## as horizontal rather than parallel to the x axis. (Note that your y component of the weight is larger than the weight w.)
 
  • Like
Likes   Reactions: aron silvester
TSny said:
The weight vector should be denoted ##\vec w##, not ##-\vec w##. The vector ##-\vec w## would point in the opposite direction of the weight.

Also, in you diagram, you drew ##\vec w_x## as horizontal rather than parallel to the x axis. (Note that your y component of the weight is larger than the weight w.)
THANKS!